Please answer a,b,c Medical researchers have developed a new artificial heart co

Question

Please answer a,b,c

Medical researchers have developed a new artificial heart constructed primarily of titanium and plastic. The heart will last and operate almost indefinitely once it is implanted in the patient's body, but the battery pack needs to be recharged about every 5 hours. A random sample of 50 battery packs is selected and subjected to a life test. The average life of these batteries is 5.04 hours. Assume that battery life is normally distributed with standard deviation sigma = .2 hours (a) Is there evidence to support the claims that mean battery life exceeds 5 hours? Use alpha = .05 and indicating 7-steps of General Procedure for Hypothesis Testing (b) What is the P-value for the test in part (a)? Does the conclusion agree with part (a)? (c) Compute the power of the test if the true mean battery life is 5 .05 hours. (d) What sample size would be require to detect a true mean battery life of 5.25 hours if we wanted the power of the test to be at least 0.9? (e) Explain how the question in part (a) could be answered by constructing a one-sided confidence bound on the mean life. (f) Show how to get the same result for part a) and b) using Minitab

Explanation / Answer

a) Null hypothesis: Ho: µ = 5
Alternative hypothesis: HA: µ > 5

= 0.05, = 0.2 hours, n = 50, xbar = 5.04 hours

One tailed z test

z = (xbar - µ)//sqrt(n) = (5.04-5)/(0.2/sqrt(50)) = 1.41421
The critical value for a right-tailed test is zc=1.64.
As z < zc, there is no sufficient evidence to reject the null hypothesis.

b) p = 0.07927, As p > 0.05, it agrees with part a and there is no sufficient evidence to reject the null hypothesis.

c) For = 0.05, Reject H0 if z >= 1.645.
z = (xbar - µ)//sqrt(n) = (xbar-5)/(0.2/sqrt(50)) > 1.645, xbar > 5.046528

If the true mean is 5.05, = P(xbar < 5.046528 | µ = 5.05)
z = (xbar - µ)//sqrt(n) = (5.046528-5.05)/(0.2/sqrt(50)) = -0.122754
= P(z < -0.122754) = 0.4512

Thus power = 1 - = 1 - 0.4512 = 0.5488 = 54.88%

Hence if true mean is 5.05, with a sample size of 50, there is only a 54.88% chance that null hypothesis will be rejected when mean is 5

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